smc

The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of $n$'s ends at position $1+2+\cdots +n$. Therefore we want to find the smallest integer $n$ that satisfies $\frac{n(n+1)}{2}\ge 1993$. By trial and error, the value of $n$ is $63$, and $63÷5$ has a remainder of $3$. $\boxed{3}$